Question

From means value theorem $f\left(b\right)-f\left(a\right)=(b-a)f^{\prime }\left(x_1\right);0<a<x_1<b$ if $f\left(x\right)=\frac{1}{x}$, then $x_1=$

• A. $\sqrt{ab}$
• B. $\frac{a+b}{2}$
• C. $\frac{2ab}{a+b}$
• D. $\frac{b-a}{b+a}$

Answer 1

The correct option is A $\sqrt{ab}$

$f\left(x\right)=1/x$

$\therefore f^{\prime }\left(x\right)=-\frac{1}{x^2}$

$\therefore f^{\prime }\left(x_1\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow \frac{-1}{x_1^2}=\frac{\frac{1}{b}-\frac{1}{a}}{(b-a)}$

$x_1=\sqrt{ab}$